When a regular polygon with n sides is decomposed into triangles using the same vertex, how many triangles are formed?(1 point)

Question

Responses

When a regular polygon is decomposed, 2⋅n triangles are formed, where n is the number of sides of the regular polygon. There are twice as many triangles formed than the number of the sides of the regular polygon.
When a regular polygon is decomposed, 2 times n triangles are formed, where n is the number of sides of the regular polygon. There are twice as many triangles formed than the number of the sides of the regular polygon.

When a regular polygon is decomposed, n triangles are formed, where n is the number of sides of the regular polygon.
When a regular polygon is decomposed, n triangles are formed, where n is the number of sides of the regular polygon.

When a regular polygon is decomposed, n−2 triangles are formed, where n is the number of sides of the regular polygon. There are two fewer triangles than the number of the sides of the regular polygon.
When a regular polygon is decomposed, n minus 2 triangles are formed, where n is the number of sides of the regular polygon. There are two fewer triangles than the number of the sides of the regular polygon.

When a regular polygon is decomposed, n+2 triangles are formed, where n is the number of sides of the regular polygon. There are two more triangles than the number of the sides of the regular polygon.

Answer 1

When a regular polygon with \( n \) sides is decomposed into triangles using the same vertex, \( n-2 \) triangles are formed.

This is because you can choose a vertex of the polygon and connect it to every other non-adjacent vertex. For a polygon with \( n \) sides, you can form \( n-2 \) triangles.

Thus, the correct response is:

"When a regular polygon is decomposed, \( n-2 \) triangles are formed, where \( n \) is the number of sides of the regular polygon. There are two fewer triangles than the number of the sides of the regular polygon."